Question: Simplify and expand the following expression: $ \dfrac{z + 3}{z - 1}+\dfrac{z - 9}{z - 7} $
Solution: In order to add expressions, they must have a common denominator. Get both fractions over a common denominator of $(z - 1)(z - 7)$ Multiply the first term by $\dfrac{z - 7}{z - 7}$ $ \begin{align*} \dfrac{z + 3}{z - 1} \times \dfrac{z - 7}{z - 7} & = \dfrac{(z + 3)(z - 7)}{(z - 1)(z - 7)} \\ & = \dfrac{z^2 - 4z - 21}{(z - 1)(z - 7)}\end{align*} $ Multiply the second term by $\dfrac{z - 1}{z - 1}$ $ \begin{align*} \dfrac{z - 9}{z - 7} \times \dfrac{z - 1}{z - 1} & = \dfrac{(z - 9)(z - 1)}{(z - 7)(z - 1)} \\ & = \dfrac{z^2 - 10z + 9}{(z - 7)(z - 1)}\end{align*} $ Now we have: $ = \dfrac{z^2 - 4z - 21}{(z - 1)(z - 7)} + \dfrac{z^2 - 10z + 9}{(z - 7)(z - 1)} $ Now both terms have a common denominator we can simply add the numerators: $ = \dfrac{z^2 - 4z - 21 + z^2 - 10z + 9}{(z - 1)(z - 7)} $ $ = \dfrac{2z^2 - 14z - 12}{(z - 1)(z - 7)}$ Expand the denominator: $ = \dfrac{2z^2 - 14z - 12}{z^2 - 8z + 7}$